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Question Number 116323 by bobhans last updated on 03/Oct/20

$$\left(1\right)\mathrm{Let}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{real}\mathrm{number}\mathrm{such}\mathrm{that}$$$$\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}=\frac{1}{3},\frac{\mathrm{bc}}{\mathrm{b}+\mathrm{c}}=\frac{1}{4}\mathrm{and}\frac{\mathrm{ac}}{\mathrm{a}+\mathrm{c}}=\frac{1}{5}.\mathrm{Find}$$$$\mathrm{the}\mathrm{value}\mathrm{of}\frac{24\mathrm{abc}}{\mathrm{ab}+\mathrm{ac}+\mathrm{bc}}?$$$$\left(2\right)\mathrm{Let}\mathrm{p}\mathrm{and}\mathrm{q}\mathrm{be}\mathrm{two}\mathrm{real}\mathrm{number}\mathrm{that}$$$$\mathrm{satisfy}\mathrm{p}.\mathrm{q}=2013.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{minimum}$$$$\mathrm{value}\mathrm{of}{(\mathrm{p}+\mathrm{q})}^2?$$

Answered by john santu last updated on 03/Oct/20

$$\left(1\right)\to \{\begin{array}{l}\frac{1}{a}+\frac{1}{b}=3\\ \frac{1}{b}+\frac{1}{c}=4\\ \frac{1}{a}+\frac{1}{c}=5\end{array}$$$$summingthethreeequation$$$$\Rightarrow 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=12$$$$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6;\frac{ab+ac+bc}{abc}=6$$$$\Rightarrow \frac{abc}{ab+ac+bc}=\frac{1}{6},multiplyboth$$$$sidesby24weget\frac{24abc}{ab+ac+bc}=4$$$$$$

Answered by john santu last updated on 03/Oct/20

$$\left(2\right){(p+q)}^2={(p-q)}^2+4pq⩾0+4\times 2013$$$$⩾8052,whenp=q=\sqrt{2013}$$$$Minimumvalueof{(p+q)}^{2}equalto8052$$

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